Difference between revisions of "2011 AIME I Problems/Problem 6"

Line 1: Line 1:
If the vertex is at (1/4, -9/8), the equation of the parabola can be expressed in the form y=a(x-1/4)^2-9/8.
+
If the vertex is at <math>(\frac{1}{4}, -\frac{9}{8})</math>, the equation of the parabola can be expressed in the form <math>y=a(x-\frac{1}{4})^2-\frac{9}{8}</math>.
Expanding, we find that y=a(x^2-x/2+1/16)-9/8 , and y=ax^2-ax/2+a/16-9/8. From the problem, we know that the parabola can be expressed in the form y=ax^2+bx+c, where a+b+c= an integer. From the above equation, we can conclude that a=a, -a/2=b, and a/16-9/8=c. Adding up all of these gives us that (9a-18)/16=a+b+c. We know that a+b+c is an integer, so 9a-18 must be divisible by 16.  Let 9a=z. If z-18=0(mod 16), z=2(mod 16). Therefore, if 9a=2, a=2/9. Adding up gives us 2+9='''11 Ans.'''
+
Expanding, we find that <math>y=a(x^2-\frac{x}{2}+\frac{1}{16})-\frac{9}{8}</math> , and <math>y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}</math>. From the problem, we know that the parabola can be expressed in the form <math>y=ax^2+bx+c</math>, where <math>a+b+c</math> is an integer. From the above equation, we can conclude that <math>a=a</math>, <math>-\frac{a}{2}=b</math>, and <math>\frac{a}{16}-\frac{9}{8}=c</math>. Adding up all of these gives us <math>\frac{9a-18}{16}=a+b+c</math>. We know that <math>a+b+c</math> is an integer, so 9a-18 must be divisible by 16.  Let <math>9a=z</math>. If <math>{z-18}\equiv {0} \pmod{16}</math>, then <math>{z}\equiv {2} \pmod{16}</math>. Therefore, if <math>9a=2</math>, <math>a=\frac{2}{9}</math>. Adding up gives us <math>2+9=\boxed{011}</math>

Revision as of 18:39, 18 March 2011

If the vertex is at $(\frac{1}{4}, -\frac{9}{8})$, the equation of the parabola can be expressed in the form $y=a(x-\frac{1}{4})^2-\frac{9}{8}$. Expanding, we find that $y=a(x^2-\frac{x}{2}+\frac{1}{16})-\frac{9}{8}$ , and $y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}$. From the problem, we know that the parabola can be expressed in the form $y=ax^2+bx+c$, where $a+b+c$ is an integer. From the above equation, we can conclude that $a=a$, $-\frac{a}{2}=b$, and $\frac{a}{16}-\frac{9}{8}=c$. Adding up all of these gives us $\frac{9a-18}{16}=a+b+c$. We know that $a+b+c$ is an integer, so 9a-18 must be divisible by 16. Let $9a=z$. If ${z-18}\equiv {0} \pmod{16}$, then ${z}\equiv {2} \pmod{16}$. Therefore, if $9a=2$, $a=\frac{2}{9}$. Adding up gives us $2+9=\boxed{011}$

Invalid username
Login to AoPS